A compass needle placed at a distance $$r$$ from a short magnet in $$\operatorname{Tan} A$$   position shows a deflection of $${60^ \circ }.$$  If the distance is increased to $$r{\left( 3 \right)^{\frac{1}{3}}},$$   then deflection of compass needle is

Solution :
$$\eqalign{ & \frac{{\tan {\theta _2}}}{{\tan {\theta _1}}} = \frac{{d_1^3}}{{\;d_2^3}} = \frac{{{r^3}}}{{{{\left[ {r{{\left( 3 \right)}^{\frac{1}{3}}}} \right]}^3}}} = \frac{1}{3} \cr & \tan {\theta _2} = \frac{1}{3}\tan {\theta _1} = \frac{{\tan 60}}{3} = \frac{{\sqrt 3 }}{3} = \frac{1}{{\sqrt 3 }} \cr & \therefore {\theta _2} = {30^ \circ } \cr} $$